Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $x = \dfrac{3t + 15}{t^2 + t - 20} \times \dfrac{t^2 - 4t}{-4t - 12} $
First factor the quadratic. $x = \dfrac{3t + 15}{(t - 4)(t + 5)} \times \dfrac{t^2 - 4t}{-4t - 12} $ Then factor out any other terms. $x = \dfrac{3(t + 5)}{(t - 4)(t + 5)} \times \dfrac{t(t - 4)}{-4(t + 3)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ 3(t + 5) \times t(t - 4) } { (t - 4)(t + 5) \times -4(t + 3) } $ $x = \dfrac{ 3t(t + 5)(t - 4)}{ -4(t - 4)(t + 5)(t + 3)} $ Notice that $(t + 5)$ and $(t - 4)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ 3t(t + 5)\cancel{(t - 4)}}{ -4\cancel{(t - 4)}(t + 5)(t + 3)} $ We are dividing by $t - 4$ , so $t - 4 \neq 0$ Therefore, $t \neq 4$ $x = \dfrac{ 3t\cancel{(t + 5)}\cancel{(t - 4)}}{ -4\cancel{(t - 4)}\cancel{(t + 5)}(t + 3)} $ We are dividing by $t + 5$ , so $t + 5 \neq 0$ Therefore, $t \neq -5$ $x = \dfrac{3t}{-4(t + 3)} $ $x = \dfrac{-3t}{4(t + 3)} ; \space t \neq 4 ; \space t \neq -5 $